Removing a compact chart disc gives a Greenian surface

The exterior is an open Riemann surface. The closed disc $\bar{B}(a,\rho )\subseteq \mathbb{C}$ is compact [F10], and ${\phi }^{-1}$ is continuous on it because $\bar{B}(a,\rho )\subseteq \phi (W)$; hence $\bar{U}={\phi }^{-1}(\bar{B}(a,\rho ))$ is compact in $X$ [F10]. Since $X$ is Hausdorff, $\bar{U}$ is closed in $X$ [F10], so $Y=X\setminus \bar{U}$ is open in $X$ and nonempty by hypothesis. The restrictions $\phi {\mathrm{\mid }}_{W\cap Y}$ of the charts of the atlas of $X$ to their (open) intersections with $Y$ form an atlas of pairwise compatible charts on $Y$: each is a homeomorphism onto an open subset of $\mathbb{C}$, and compatibility is inherited from $X$. With this atlas $Y$ satisfies all clauses of [F1]: it is Hausdorff and second countable as a subspace of $X$, and its connectedness is proved separately below; the complex structure it carries is the one induced by $X$.

The boundary circle is compact and connected. Put $\gamma :=\mathrm{\partial }U={\phi }^{-1}(\mathrm{\partial }B(a,\rho ))$. It is nonempty, and it is compact as the continuous image of the compact circle $\mathrm{\partial }B(a,\rho )$ [F10]; it is also connected, being the image of the connected space $\mathbb{R}$ under the continuous map $t\mapsto {\phi }^{-1}(a+\rho e^{it})$ [F11]. The circle lies in $\bar{U}$ and is disjoint from $Y=X\setminus \bar{U}$.

Chartwise strong maximum principle. Let $V$ be a nonempty connected open subset of a Riemann surface and let $u$ be subharmonic on $V$. If $u$ attains a finite maximum $S$ at some point $x\in V$, then $u\equiv S$ on $V$. Indeed, choose a chart $\psi :V^{\mathrm{\prime }}\to \mathbb{C}$ of $V$ with $x\in V^{\mathrm{\prime }}$ [F1]; the chart expression $u_{\psi }$ is plane subharmonic on the plane domain $\psi (V^{\mathrm{\prime }})$ [F5] and attains the finite maximum $S$ at $\psi (x)$, so $u_{\psi }\equiv S$ on $\psi (V^{\mathrm{\prime }})$ by [F7]; hence the set $Z:=\{u=S\}$ is open, and it is closed in $V$ because $u$ is upper semicontinuous [F5, F9]; as $V$ is connected and $Z\mathrm{\ne }\mathrm{\varnothing }$, $Z=V$.

A pole disc with an extending coordinate. Fix $p_0\in Y$. Restrict a chart centred at $p_0$ to a small Euclidean disc whose closure lies inside its original chart image; scale its coordinate so that this restricted disc is $U_1=\{\mathrm{\mid }z\mathrm{\mid }<1\}$ and the same coordinate $z$ is defined on a larger neighbourhood of ${\bar{U}}_1$ in $Y$. Thus ${\bar{U}}_1$ is compact in $Y$. Fix $0<r<1$ and put $r{U}_1=\{\mathrm{\mid }z\mathrm{\mid }<r\}$; the circles $\mathrm{\partial }(r{U}_1)=\{\mathrm{\mid }z\mathrm{\mid }=r\}$ and $\mathrm{\partial }{U}_1=\{\mathrm{\mid }z\mathrm{\mid }=1\}$ are now legitimate coordinate circles.

$Y$ is connected. Suppose $Y=A\bigsqcup B$ with $A,B$ nonempty, open in $Y$; since $Y$ is open in $X$ [step 1.1], both $A$ and $B$ are open in $X$. The closed disc satisfies $\bar{U}=U\bigsqcup \gamma$, so $X=U\bigsqcup \gamma \bigsqcup A\bigsqcup B$ is a disjoint union. Every point $x\in \gamma$ lies in the closure of $A\cup B$: a neighbourhood of $x$ of the form ${\phi }^{-1}(B(\phi (x),\epsilon ))$ contains points of $U={\phi }^{-1}(B(a,\rho ))$ and points of ${\phi }^{-1}(\mathbb{C}\setminus \bar{B}(a,\rho ))\subseteq Y=A\cup B$, because $\phi (x)\in \mathrm{\partial }B(a,\rho )$ and every disc around a boundary point of $B(a,\rho )$ meets both the open disc and its exterior. Hence $\gamma \subseteq \bar{A}\cup \bar{B}$. These two traces on $\gamma$ are disjoint: if $x$ lay in both, a sufficiently small chart neighbourhood of $x$ would have a connected exterior half-disc contained in $Y$ and meeting both $A$ and $B$, contrary to the separation $Y=A\bigsqcup B$. Since $\gamma$ is connected [step 1.2], the two disjoint closed traces covering it force $\gamma \subseteq \bar{A}$ or $\gamma \subseteq \bar{B}$. In the first case the open set $B$ satisfies $B=X\setminus \overline{A\cup U}$: the inclusion $B\subseteq X\setminus \overline{A\cup U}$ holds because $B$ is open and disjoint from $A\cup U$; conversely if $x\mathrm{\notin }\overline{A\cup U}$ then $x\mathrm{\notin }A\cup U$, while $x\in \gamma$ is impossible because $\gamma \subseteq \bar{A}\subseteq \overline{A\cup U}$, so $x\in B$ by the disjoint decomposition of $X$. Moreover $\bar{B}\cap \gamma =\mathrm{\varnothing }$ by the disjoint-trace assertion, and $\bar{B}$ meets neither $A$ nor $U$, since these are open and disjoint from $B$. Hence $\bar{B}=B$. Thus $B$ is both open and closed in the connected space $X$ [F1] and $B\mathrm{\ne }\mathrm{\varnothing }$, so $B=X$ and $A=\mathrm{\varnothing }$, a contradiction. The case $\gamma \subseteq \bar{B}$ is symmetric, interchanging $A$ and $B$. Therefore $Y$ is connected, and by step 1.1 it is a Riemann surface.

Boundary maximum principle on a relatively compact domain. Let $V\subseteq X$ be a nonempty proper open connected subset with $\bar{V}$ compact, and let $u:V\to [-\mathrm{\infty },\mathrm{\infty })$ be subharmonic on $V$ with ${\mathrm{lim\hspace{0.17em}sup}}_{q\to \zeta ,q\in V}u(q)\le 0$ for every $\zeta \in \mathrm{\partial }V$. Then $u\le 0$ on $V$. For each $b>0$, the set $E_b=\{u\ge b\}$ is closed in $\bar{V}$ and avoids $\mathrm{\partial }V$ by the boundary limsup hypothesis; it is compact. If ${\sup }_{V}u=+\mathrm{\infty }$, the nested nonempty $E_n$ have nonempty intersection by compactness, giving a forbidden $+\mathrm{\infty }$ value. If $S:={\sup }_{V}u$ is finite and positive, the nested nonempty $E_b$ for $0<b<S$ give an interior point with value $S$. Step 1.3 forces $u\equiv S$, contradicting the boundary limsup bound at any point of the nonempty $\mathrm{\partial }V$. Thus $u\le 0$.

Maximum principle with boundary values on a compactly contained disc. Let $V$ be a nonempty proper open connected subset of $X$ with $\bar{V}$ compact, let $u:\bar{V}\to [-\mathrm{\infty },\mathrm{\infty })$ be upper semicontinuous on $\bar{V}$ and subharmonic on $V$. Then ${\sup }_{\bar{V}}u={\max }_{\mathrm{\partial }V}u$. Upper semicontinuity bounds $u$ above on compact $\bar{V}$: the open strict sublevel sets $\{u<n\}$ cover it, so a finite subcover bounds $u$. It has a finite value somewhere in $V$, as it is subharmonic there. Thus $S:={\sup }_{\bar{V}}u$ is real, and the nonempty closed superlevel sets $\{u\ge b\}$, $b<S$, have the finite-intersection property on $\bar{V}$. Compactness gives a point with value $S$ [F10]. If a maximiser $x$ lies in $\mathrm{\partial }V$, then $S={\max }_{\mathrm{\partial }V}u$. If every maximiser lies in $V$, then $u\equiv S$ on $V$ by step 1.3, and for $\zeta \in \mathrm{\partial }V$ upper semicontinuity at $\zeta$ (which lies in the domain of $u$) gives $u(\zeta )\ge {\mathrm{lim\hspace{0.17em}sup}}_{q\to \zeta }u(q)\ge S$; hence $S\le {\max }_{\mathrm{\partial }V}u$. In both cases $S\le {\max }_{\mathrm{\partial }V}u$, and the reverse inequality is trivial.

Inequality (5) at the pole. Let $v\in {\mathcal{F}}_{p_0}(Y)$ and $\epsilon >0$, and put $w:=v+(1+\epsilon )\log \mathrm{\mid }z\mathrm{\mid }$ on $U_1\setminus \{p_0\}$. Then $w$ extends to a function on ${\bar{U}}_1$ that is upper semicontinuous there and subharmonic on $U_1$: on $U_1\setminus \{p_0\}$ the chart expression of $v$ is plane subharmonic [F5], the function $\log \mathrm{\mid }\cdot \mathrm{\mid }$ is harmonic on $\mathbb{D}\setminus \{0\}$ [F13], so $w$ is subharmonic on $U_1\setminus \{p_0\}$ chartwise [F5, F8, F14]; and the pole condition of [F2] gives $v_z(\zeta )\le -\log \mathrm{\mid }\zeta \mathrm{\mid }+C$ near $0$, whence $w_z(\zeta )\le \epsilon \log \mathrm{\mid }\zeta \mathrm{\mid }+C\to -\mathrm{\infty }$ as $\zeta \to 0$, so setting $w(p_0):=-\mathrm{\infty }$ makes $w$ upper semicontinuous at $p_0$, and the submean inequality at $p_0$ holds trivially for the value $-\mathrm{\infty }$ [F6]; subharmonicity of this extension follows by truncation: $\max (w,-N)$ is constant near $p_0$, subharmonic elsewhere, and hence subharmonic by locality; its decreasing submean inequalities pass to $w$ by monotone convergence on circles after subtraction of a finite common upper bound. This is also the extension argument in the proof of [F3]. On $\mathrm{\partial }{U}_1$ one has $\mathrm{\mid }z\mathrm{\mid }=1$, so $w=v$ there; step 2.3 applied to $w$ on the disc $U_1$ therefore gives ${\max }_{{\bar{U}}_1}w={\max }_{\mathrm{\partial }{U}_1}w={\max }_{\mathrm{\partial }{U}_1}v=:M_v\mathrm{.}$ Restricting to the smaller circle $\mathrm{\partial }(r{U}_1)$, where $w=v+(1+\epsilon )\log r$, gives ${\max }_{\mathrm{\partial }(r{U}_1)}v+(1+\epsilon )\log r\le M_v\mathrm{.}$ Letting $\epsilon \downarrow 0$ yields

A fixed barrier and candidate-dependent truncations. Fix $q_1\in \mathrm{\partial }{U}_1$. If $X$ is compact, fix $a_0\in U$ and use the noncompact ambient surface $Z:=X\setminus \{a_0\}$ for all Dirichlet applications. It is connected: a punctured coordinate disc about $a_0$ is connected, so any separation of $Z$ would extend to one of $X$ by adjoining $a_0$ to the side containing that punctured disc. It is noncompact, since compactness would make $Z$ closed in the Hausdorff $X$, making $a_0$ isolated, contrary to a coordinate chart. Set $V_0:=X\setminus (\bar{U}\cup \overline{r{U}_1})$. If $X$ is noncompact, choose by [F4] a regular exhaustion $(D_n)$ and $N_0$ with $\bar{U}\cup {\bar{U}}_1\subseteq D_{N_0}$, and put $V_0:=D_{N_0}\setminus (\bar{U}\cup \overline{r{U}_1})$. In either case $V_0$ is a connected relatively compact smooth-bordered domain by applying the closed-disc removal argument of step 2.1 successively in the connected ambient domain. Its closure lies in $Z$ in the compact case. Solve on $V_0$ for a continuous harmonic $\eta$ with boundary values $1$ on $\mathrm{\partial }(r{U}_1)$, $0$ on $\mathrm{\partial }U$, and, when $X$ is noncompact, $1$ on $\mathrm{\partial }{D}_{N_0}$ [F4]. The maximum principle step 2.2 gives $0\le \eta \le 1$. Since $\mathrm{\partial }{U}_1$ lies in $V_0$ and $1-\eta$ is nonnegative harmonic, the strong maximum principle step 1.3 gives $1-\eta >0$ on $\mathrm{\partial }{U}_1$; hence $\delta :={\min }_{\mathrm{\partial }{U}_1}(1-\eta )>0$. Now fix any $v\in {\mathcal{F}}_{p_0}(Y)$. If $X$ is compact, set $V:=V_0$ and $\omega :=\eta$. If $X$ is noncompact, choose $N\ge N_0$ with the compact support of $v$ contained in $D_N$, set $V:=D_N\setminus (\bar{U}\cup \overline{r{U}_1})$. As with $V_0$, this is a nonempty connected relatively compact smooth-bordered domain: the two closed discs are disjoint and contained in $D_N$, so successive applications of step 2.1 give connectedness. Solve by [F4] for $\omega =1$ on $\mathrm{\partial }(r{U}_1)$ and $\omega =0$ on $\mathrm{\partial }U\cup \mathrm{\partial }{D}_N$. Then $0\le \omega \le 1$ by step 2.2. On $V_0$, $\eta -\omega$ is harmonic, vanishes on $\mathrm{\partial }(r{U}_1)\cup \mathrm{\partial }U$, and is nonnegative on $\mathrm{\partial }{D}_{N_0}$ because $\eta =1$ there and $\omega \le 1$; hence $\omega \le \eta$ on $V_0$. In both cases $0\le \omega \le 1$ on $V$ and ${\max }_{\mathrm{\partial }{U}_1}\omega \le 1-\delta$, with $\delta$ independent of $v$.

Inequality (6). Let $v\in {\mathcal{F}}_{p_0}(Y)$ and put $M_v:={\max }_{\mathrm{\partial }(r{U}_1)}v$, a nonnegative real number. The function $u:=v-M_v\omega$ is subharmonic on $V$: chartwise, $v$ is subharmonic on $V\subseteq Y\setminus \{p_0\}$ [F5], $\omega$ is harmonic on $V$ hence subharmonic with each nonnegative multiple [F14], and the chart expression of $-M_v\omega$ is the sum of the plane subharmonic function $v_{\phi }$ and the plane subharmonic function $-M_v{\omega }_{\phi }$, hence subharmonic [F8, F5]. At every boundary point $\zeta$ of $V$ the boundary limit of $u$ is at most $0$: for $\zeta \in \mathrm{\partial }(r{U}_1)$ one has $\omega (\zeta )=1$ and $v$ is upper semicontinuous at $\zeta$ [F5, F9], so $\mathrm{lim\hspace{0.17em}sup}(v-M_v\omega )\le v(\zeta )-M_v\le 0$; for $\zeta \in \mathrm{\partial }U$ the point $\zeta$ has a neighbourhood disjoint from the compact support of $v$ [F2], so $v=0$ near $\zeta$ and $\omega (\zeta )=0$, giving $\mathrm{lim\hspace{0.17em}sup}u\le 0$; and in the noncompact case the same argument applies at $\zeta \in \mathrm{\partial }{D}_N$, because $\bar{U}\cup {\bar{U}}_1\subseteq D_N$ while $v$ vanishes off a compact subset of $Y$, so $v=0$ on a neighbourhood of $\mathrm{\partial }{D}_N$. Step 2.2 applied on $V$ therefore gives $v-M_v\omega \le 0$ on $V$, and in particular on $\mathrm{\partial }{U}_1\subseteq V$:

The family is uniformly bounded at the pole. Adding (5) and (6) gives ${\max }_{\mathrm{\partial }(r{U}_1)}v+\log r\le (1-\delta ){\max }_{\mathrm{\partial }(r{U}_1)}v,\text{hence}\delta {\max }_{\mathrm{\partial }(r{U}_1)}v\le \log \frac{1}{r},$ so that for every $v\in {\mathcal{F}}_{p_0}(Y)$ ${\max }_{\mathrm{\partial }(r{U}_1)}v\le C:=\frac{1}{\delta }\log \frac{1}{r}<+\mathrm{\infty }\mathrm{.}$

One finite value of the envelope. For $v\in {\mathcal{F}}_{p_0}(Y)$, step 4.1 gives $v(q_1)\le M_v\omega (q_1)\le C$ with $C$ as in step 5.1, because $q_1\in \mathrm{\partial }{U}_1\subseteq V$ and $0\le \omega \le 1$; taking the supremum over $v$, $g_Y(q_1,p_0)=\sup \{v(q_1):v\in {\mathcal{F}}_{p_0}(Y)\}\le C<+\mathrm{\infty }\mathrm{.}$

Conclusion. Since the envelope of ${\mathcal{F}}_{p_0}(Y)$ is finite at the point $q_1\in Y\setminus \{p_0\}$, the dichotomy [F3] shows that it is finite everywhere on $Y\setminus \{p_0\}$, harmonic and positive there, and has a unit logarithmic pole at $p_0$; that is, $Y$ admits a finite canonical Green kernel at $p_0$. The pole $p_0\in Y$ was arbitrary, so $Y$ admits a finite canonical Green kernel at every pole, and $Y$ is connected by step 2.1 and a Riemann surface by step 1.1. The countably many arbitrary choices of the construction are the exhausted domains of [F4] in the noncompact case, which uses ${\mathrm{A}\mathrm{C}}_{\omega }$ [A1]; all other selections are finite.